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Chapter 15. Applying equilibrium. The Common Ion Effect. When the salt with the anion of a weak acid is added to that acid, Lowers the percent dissociation of the acid. Ex: NaF and HF mixed together F- is the common ion HF ↔ H + + F -

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chapter 15

Chapter 15

Applying equilibrium

the common ion effect
The Common Ion Effect
  • When the salt with the anion of a weak acid is added to that acid,
  • Lowers the percent dissociation of the acid.
  • Ex: NaF and HF mixed together
    • F- is the common ion

HF ↔ H+ + F-

Adding F- shifts the equilibrium to the left and therefore fewer H+ ions present

slide3

The same principle applies to salts with the cation of a weak base.

The calculations are the same as last chapter.

NH4Cl and NH3 (NH4+ is the common ion)

NH3 + H+ NH4+

buffered solutions
Buffered solutions
  • A solution that resists a change in pH.
  • Consists of either a weak acid and its salt or a weak base and its salt.
  • We can make a buffer of any pH by varying the concentrations of these solutions.
  • Same calculations as before.
  • Calculate the pH of a solution that is .50 M HAc and .25 M NaAc (Ka = 1.8 x 10-5)
slide5

Na+ is a spectator and the reaction we are worried about is

HAc H+ + Ac-

  • Choose x to be small

Initial 0.50 M 0 0.25 M

-x

x

x

Change

Final

0.50-x

x

0.25+x

  • We can fill in the table
slide6

HAc H+ + Ac-

Initial 0.50 M 0 0.25 M

  • Do the math
  • Ka = 1.8 x 10-5

Change

-x

x

x

Final

0.50-x

x

0.25+x

x (0.25)

x (0.25+x)

=

1.8 x 10-5 =

(0.50)

(0.50-x)

  • Assume x is small

x = 3.6 x 10-5

  • Assumption is valid
  • pH = -log (3.6 x 10-5) = 4.44
adding a strong acid or base
Adding a strong acid or base
  • Do the stoichiometry first.
    • Use moles not molar
  • A strong base will grab protons from the weak acid reducing [HA]0
  • A strong acid will add its proton to the anion of the salt reducing [A-]0
  • Then do the equilibrium problem.
  • What is the pH of 1.0 L of the previous solution when 0.010 mol of solid NaOH is added?
slide8

HAc H+ + Ac-

0.25 mol

0.50 mol

  • In the initial mixture M x L = mol
  • 0.50 M HAc x 1.0 L = 0.50 mol HAc

0.26 mol

0.49 mol

  • 0.25 M Ac- x 1.0 L = 0.25 mol Ac-
  • Adding 0.010 mol OH- will reduce the HAc and increase the Ac- by 0.010 mole
  • Because it is in 1.0 L, we can convert it to molarity
slide9

HAc H+ + Ac-

0.25 mol

0.50 mol

0.26 M

0.49 M

  • In the initial mixture M x L = mol
  • 0.50 M HAc x 1.0 L = 0.50 mol HAc
  • 0.25 M Ac- x 1.0 L = 0.25 mol Ac-
  • Adding 0.010 mol OH- will reduce the HAc and increase the Ac- by 0.010 mole
  • Because it is in 1.0 L, we can convert it to molarity
slide10

HAc H+ + Ac-

0.25 mol

0.50 mol

0.26 M

0.49 M

  • Fill in the table

HAc H+ + Ac-

Initial 0.49 M 0 0.26 M

-x

x

x

Final

0.49-x

x

0.26+x

slide11

HAc H+ + Ac-

Initial 0.49 M 0 0.26 M

  • Do the math
  • Ka = 1.8 x 10-5

-x

x

x

Final

0.49-x

x

0.26+x

x (0.26)

x (0.26+x)

=

1.8 x 10-5 =

(0.49)

(0.49-x)

  • Assume x is small

x = 3.4 x 10-5

  • Assumption is valid
  • pH = -log (3.4 x 10-5) = 4.47
notice
Notice
  • If we had added 0.010 mol of NaOH to 1 L of water, the pH would have been.
  • 0.010 M OH-
  • pOH = 2
  • pH = 12
  • But with a mixture of an acid and its conjugate base the pH doesn’t change much
  • Called a buffer.
general equation
General equation
  • Ka = [H+] [A-] [HA]
  • so [H+] = Ka [HA] [A-]
  • The [H+] depends on the ratio [HA]/[A-]
  • taking the negative log of both sides
  • pH = -log(Ka [HA]/[A-])
  • pH = -log(Ka)-log([HA]/[A-])
  • pH = pKa + log([A-]/[HA])
this is called the henderson hasselbach equation
This is called the Henderson-Hasselbach equation
  • pH = pKa + log([A-]/[HA])
  • pH = pKa + log(base/acid)
  • Works for an acid and its salt
  • Like HNO2 and NaNO2
  • Or a base and its salt
  • Like NH3 and NH4Cl
  • But remember to change Kb to Ka
slide15
Calculate the pH of the following
  • 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4)
  • pH = 3.38
slide16
Calculate the pH of the following
  • 0.25 M NH3 and 0.40 M NH4Cl (Kb = 1.8 x 10-5)
  • Ka = 1 x 10-14 1.8 x 10-5
  • Ka = 5.6 x 10-10
  • remember its the ratio base over acid
  • pH = 9.05
prove they re buffers
Prove they’re buffers
  • What would the pH be if .020 mol of HCl is added to 1.0 L of both of the preceding solutions.
  • What would the pH be if 0.050 mol of solid NaOH is added to 1.0 L of each of the proceeding.
  • Remember adding acids increases the acid side,
  • Adding base increases the base side.
prove they re buffers1
Prove they’re buffers
  • What would the pH be if .020 mol of HCl is added to 1.0 L of preceding solutions.
  • 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4)

HC3H5O3 H+ + C3H5O3-

Initially 0.75 mol 0 0.25 mol

After acid 0.77 mol 0.23 mol

Compared to 3.38 before acid was added

prove they re buffers2
Prove they’re buffers
  • What would the pH be if 0.050 mol of solid NaOH is added to 1.0 L of the solutions.
  • 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4)

HC3H5O3 H+ + C3H5O3-

Initially 0.75 mol 0 0.25 mol

After acid 0.70 mol 0.30 mol

Compared to 3.38 before acid was added

prove they re buffers3
Prove they’re buffers
  • What would the pH be if .020 mol of HCl is added to 1.0 L of preceding solutions.
  • 0.25 M NH3 and 0.40 M NH4Cl Kb = 1.8 x 10-5 so Ka = 5.6 x 10-10

NH4+H+ + NH3

Initially 0.40 mol 0 0.25 mol

After acid 0.42 mol 0.23 mol

Compared to 9.05 before acid was added

prove they re buffers4
Prove they’re buffers
  • What would the pH be if 0.050 mol of solid NaOH is added to 1.0 L each solutions.
  • 0.25 M NH3 and 0.40 M NH4Cl Kb = 1.8 x 10-5 so Ka = 5.6 x 10-10

NH4+H+ + NH3

Initially 0.40 mol 0 0.25 mol

After acid 0.35 mol 0.30 mol

Compared to 9.05 before acid was added

buffer capacity
Buffer capacity
  • The pH of a buffered solution is determined by the ratio [A-]/[HA].
  • As long as this doesn’t change much the pH won’t change much.
  • The more concentrated these two are the more H+ and OH- the solution will be able to absorb.
  • Larger concentrations = bigger buffer capacity.
buffer capacity1
Buffer Capacity
  • Calculate the change in pH that occurs when 0.040 mol of HCl(g) is added to 1.0 L of each of the following:
  • 5.00 M HAc and 5.00 M NaAc
  • 0.050 M HAc and 0.050 M NaAc
  • Ka= 1.8x10-5
  • pH = pKa
buffer capacity2
Buffer Capacity
  • Calculate the change in pH that occurs when 0.040 mol of HCl(g) is added to 1.0 L of 5.00 M HAc and 5.00 M NaAc
  • Ka= 1.8x10-5

HAc H+ + Ac-

Initially 5.00 mol 0 5.00 mol

After acid 5.04 mol 4.96 mol

Compared to 4.74 before acid was added

buffer capacity3
Buffer Capacity
  • Calculate the change in pH that occurs when 0.040 mol of HCl(g) is added to 1.0 L of 0.050 M HAc and 0.050 M NaAc
  • Ka= 1.8x10-5

HAc H+ + Ac-

Initially 0.050 mol 0 0.050 mol

After acid 0.090 mol 0 0.010 mol

Compared to 4.74 before acid was added

buffer capacity4
Buffer capacity
  • The best buffers have a ratio [A-]/[HA] = 1
  • This is most resistant to change
  • True when [A-] = [HA]
  • Makes pH = pKa (since log 1 = 0)
titrations1
Titrations
  • Millimole (mmol) = 1/1000 mol
  • Molarity = mmol/mL = mol/L
  • Makes calculations easier because we will rarely add liters of solution.
  • Adding a solution of known concentration until the substance being tested is consumed.
  • This is called the equivalence point.
  • Where moles of acid = moles of base
slide29
Strong acid with strong Base
  • Equivalence at pH 7

7

pH

mL of Base added

slide30

Weak acid with strong Base

  • Equivalence at pH >7

>7

7

pH

mL of Base added

slide31

Strong base with strong acid

  • Equivalence at pH 7

7

pH

mL of acid added

slide32

Weak base with strong acid

  • Equivalence at pH <7

7

<7

pH

mL of acid added

strong acid with strong base
Strong acid with Strong Base
  • Do the stoichiometry.
  • mL x M = mmol
  • There is no equilibrium .
  • They both dissociate completely.
  • The reaction is H+ + OH- HOH
  • Use [H+] or [OH-] to figure pH or pOH
  • The titration of 20.0 mL of 0.10 M HNO3 with 0.10 M NaOH
weak acid with strong base
Weak acid with Strong base
  • There is an equilibrium.
  • Do stoichiometry.
    • Use moles
  • Determine major species
  • Then do equilibrium.

Calculate the pH after 0.0mL, 15.0mL, 25.0 mL, and 30.0mL of 0.10 M NaOH are added to 25.0mL of 0.10M of Nicotinic acid Ka= 1.4 x 10 -5

summary
Summary
  • Strong acid and base just stoichiometry.
  • Weak acid with 0 ml of base - Ka
  • Weak acid before equivalence point
      • Stoichiometry first
      • Then Henderson-Hasselbach
  • Weak acid at equivalence point- Kb -Calculate concentration
  • Weak acid after equivalence - leftover strong base. -Calculate concentration
example
Example
  • A 25.0 mL sample of benzoic acid (0.120M) is titrated w/ NaOH (.210 M) Ka= 6.3 x 10 -5
    • Calculate # moles of Benzoic acid
    • Calculate volume of NaOH needed to reach equiv. pt
    • Calc. the pH before any base is added
    • Calc. pH after 10ml base is added
    • Find the pH at equivalence pt
    • Calculate the pH after 15.3mL of OH- is added
summary1
Summary
  • Weak base before equivalence point.
      • Stoichiometry first
      • Then Henderson-Hasselbach
  • Weak base at equivalence point Ka. -Calculate concentration
  • Weak base after equivalence – left over strong acid. -Calculate concentration
indicators
Indicators
  • Weak acids that change color when they become bases.
  • weak acid written HIn
  • Weak base
  • HIn H+ + In-clear red
  • Equilibrium is controlled by pH
  • End point - when the indicator changes color.
  • Try to match the equivalence point
indicators1
Indicators
  • Since it is an equilibrium the color change is gradual.
  • It is noticeable when the ratio of [In-]/[HI] is 1/10 or [HI]/[In-] is 10/1
  • Since the Indicator is a weak acid, it has a Ka.
  • pH the indicator changes is
  • pH=pKa +log([In-]/[HI]) = pKa +log(1/10)
  • pH=pKa - 1
indicators2
Indicators
  • pH=pKa + log([HI]/[In-]) = pKa + log(10)
  • pH=pKa+1
  • Choose the indicator with a pKa 1 more than the pH at equivalence point if you are titrating with base.
  • Choose the indicator with a pKa 1 less than the pH at equivalence point if you are titrating with acid.
sample problem
Sample Problem
  • Two drops of indicator HIn (Ka= 1.0 x 10 -9), hwere HIn is yellow and In- is blue are placed in 100.0 mL of 0.10 M HCl.
    • What color is the solution initially?
    • The solution is titrated with 0.10 M NaOH. At what pH will the color change (greenish yellow) occur?
    • What color will the solution be after 200.0mL of NaOH has been added?
solubility equilibria

Solubility Equilibria

Will it all dissolve, and if not, how much?

slide45
All dissolving is an equilibrium.
  • If there is not much solid it will all dissolve.
  • As more solid is added the solution will become saturated.
  • Solid dissolved
  • The solid will precipitate as fast as it dissolves .
  • Equilibrium
general equation1
General equation
  • M+ stands for the cation (usually metal).
  • Nm-stands for the anion (a nonmetal).
  • MaNmb(s) aM+(aq) + bNm- (aq)
  • Remember the concentration of a solid doesn’t change. So, what’s the K expression?
  • Ksp = [M+]a[Nm-]b
  • Called the solubility product for each compound.
watch out
Watch out
  • Solubility is not the same as solubility product.
  • Solubility product is an equilibrium constant.
  • it doesn’t change except with temperature.
  • Solubility is an equilibrium position-usually expressed as a Molarity for how much can dissolve.
calculating solubility
Calculating Solubility

Ex: Calc. Solubility of CaC2O4 if the value of Ksp= 4.8 x 10 -5 mol/L

CaC2O4 (s) ↔ Ca+2(aq) + C2O4 -2 (aq)

Ksp = [Ca+2] [C2O4-2]

Ksp= [s] [s] = [s]2 = 4.8x10-5 = s2

s= 6.9 x 10 -3 M

calculating k sp
Calculating Ksp
  • The solubility of copper(I) bromide is

2.0 x 10 -4 mol/L. Calc. Ksp value.

CuBr(s) ↔ Cu+1(aq) + Br-1 (aq)

I .0002M 0M 0M

C - .0002M +.0002M +.0002M

E 0 .0002M .0002M

Ksp = [.0002] [.0002] = 4.0 x 10 -8

(units usually omitted)

calculating k sp1
Calculating Ksp
  • The solubility of iron(II) oxalate FeC2O4 is 65.9 mg/L
  • The solubility of Li2CO3 is 5.48 g/L
common ion effect
Common Ion Effect
  • If we try to dissolve the solid in a solution with either the cation or anion already present less will dissolve.
  • Calculate the solubility of SrSO4, with a Ksp of 3.2 x 10-7 in M in a solution of 0.010 M Na2SO4.
  • Calculate the solubility of CaF2

Ksp= 4.0 x 10 -11 in a 0.025 M NaF solution.

agcl and nacl in solution
AgCl and NaCl in solution
  • Cl- is the common ion
  • When [Ag+] [Cl-] < Ksp, no precipitate will be formed.
  • When [Ag+] [Cl-] > Ksp, a precipitate will be formed.
ph and solubility
pH and solubility
  • OH- can be a common ion.
  • More soluble in acid.
  • For other anions if they come from a weak acid they are more soluble in a acidic solution than in water.
  • CaC2O4 Ca+2 + C2O4-2
  • H+ + C2O4-2 HC2O4-
  • Reduces [C2O4-2] in acidic solution.
sample problem1
Sample problem
  • What is the pH in a saturated solution of Ca(OH)2?Ksp = 5.5 x 10-6 for Ca(OH)2.
q and precipitation
Q and Precipitation
  • not necessarily at Equil.
  • Q= Ion product
  • Q =[M+]a[Nm-]b
  • If Q>Ksp a precipitate forms. (supersaturated)
  • If Q<Ksp No precipitate. (unsaturated)
  • If Q = Ksp equilibrium. (saturated)
example1
Example1

Chemical analysis gave [Pb2+] = 0.012 M, and [Br-] = 0.024 M in a solution. From a table, you find Ksp for PbBr2 has a value of 4x 10-5. Is the solution saturated, oversaturated or unsaturated?

example 2
Example 2
  • A solution of 750.0 mL of 4.00 x 10-3M Ce(NO3)3 is added to 300.0 mL of 2.00 x 10-2M KIO3. Will Ce(IO3)3 (Ksp= 1.9 x 10-10M) precipitate and if so, what is the concentration of the ions?
selective precipitations
Selective Precipitations
  • Used to separate mixtures of metal ions in solutions.
  • Add anions that will only precipitate certain metals at a time.
  • Used to purify mixtures.
  • Often use H2S because in acidic solution Hg+2, Cd+2, Bi+3, Cu+2, Sn+4 will precipitate.
selective precipitation
Selective Precipitation
  • Then add OH-solution [S-2] will increase so more soluble sulfides will precipitate.
  • Co+2, Zn+2, Mn+2, Ni+2, Fe+2, Cr(OH)3, Al(OH)3
selective precipitation1
Selective precipitation
  • Follow the steps
  • First with insoluble chlorides (Ag, Pb, Ba)
  • Then sulfides in Acid.
  • Then sulfides in base.
  • Then insoluble carbonate (Ca, Ba, Mg)
  • Alkali metals and NH4+ remain in solution.
complex ion equilibria
Complex ion Equilibria
  • A charged ion surrounded by ligands.
  • Ligands are Lewis bases using their lone pair to stabilize the charged metal ions.
  • Common ligands are NH3, H2O, Cl-,CN-
  • Coordination number is the number of attached ligands.
  • Cu(NH3)42+ has a coordination # of 4
the addition of each ligand has its own equilibrium
The addition of each ligand has its own equilibrium
  • Usually the ligand is in large excess.
  • And the individual K’s will be large so we can treat them as if they go to completion.
  • The complex ion will be the biggest ion in solution.
slide64
Calculate the concentrations of Ag+, Ag(S2O3)-2, and Ag(S2O3)2-3 in a solution made by mixing 150.0 mL of 0.010 M AgNO3 with 200.0 mL of 5.00 M Na2S2O3
  • Ag+ + S2O3-2 Ag(S2O3)- K1=7.4 x 108
  • Ag(S2O3)- + 2S2O3-2 Ag(S2O3)2-3 K2=3.9 x 104

(Hint: set up ice problem, determine L.R.)

This problem is found on pg. 768-769 in your book. Check your work!

complex ions and solubility
Complex Ions and Solubility

The dissolving of solid AgCl in excess NH3 is represented as…

AgCl(s) + 2NH3(aq) ↔ Ag(NH3)2+ (aq) + Cl-(aq)

However, this is a result of several steps

AgCl(s) ↔ Ag+ + Cl- Ksp= 1.6 x 10-10

Ag+ + NH3 ↔ Ag(NH3)+ K1 = 2.1 x 103

Ag(NH3)+ + NH3 ↔ Ag(NH3)2+ K2= 8.2 x 103

agcl s 2nh 3 aq ag nh 3 2 aq cl aq
AgCl(s) + 2NH3(aq) ↔ Ag(NH3)2+ (aq) + Cl-(aq)

So the Equilibrium Expression is written as…. K= [Ag(NH3)2+] [Cl-]

[NH3]2

= Ksp x K1 x K2

= (1.6 x 10-10) ( 2.1 x 103) (8.2 x 103)

= 2.8 x 10-3

summary2
Summary

3 Factors that affect solubility:

- Common Ions

- pH

- Complexing Agents/Ions

Mixtures of Ions can be separated by selective precipitation (you know this as Quantitative Analysis)

- Steps listed on pg.765 or in your lab